Solve 10/2x-3=x-1

$$\frac{ 10 }{ 2x-3 } = x-1$$

Solve for x

$$x=-1$$

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x-3.

$$10=\left(2x-3\right)x+\left(2x-3\right)\left(-1\right)$$

Use the distributive property to multiply 2x-3 by x.

$$10=2x^{2}-3x+\left(2x-3\right)\left(-1\right)$$

Use the distributive property to multiply 2x-3 by -1.

$$10=2x^{2}-3x-2x+3$$

Combine -3x and -2x to get -5x.

$$10=2x^{2}-5x+3$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-5x+3=10$$

Subtract 10 from both sides.

$$2x^{2}-5x+3-10=0$$

Subtract 10 from 3 to get -7.

$$2x^{2}-5x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-7\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-7\right)}}{2\times 2}$$

Multiply -8 times -7.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\times 2}$$

Add 25 to 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-\left(-5\right)±9}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±9}{4}$$

Now solve the equation x=\frac{5±9}{4} when ± is plus. Add 5 to 9.

$$x=\frac{14}{4}$$

Reduce the fraction \frac{14}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{5±9}{4} when ± is minus. Subtract 9 from 5.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=\frac{7}{2} x=-1$$

Steps for Completing the Square

Variable x cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x-3.

$$10=\left(2x-3\right)x+\left(2x-3\right)\left(-1\right)$$

Use the distributive property to multiply 2x-3 by x.

$$10=2x^{2}-3x+\left(2x-3\right)\left(-1\right)$$

Use the distributive property to multiply 2x-3 by -1.

$$10=2x^{2}-3x-2x+3$$

Combine -3x and -2x to get -5x.

$$10=2x^{2}-5x+3$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-5x+3=10$$

Subtract 3 from both sides.

$$2x^{2}-5x=10-3$$

Subtract 3 from 10 to get 7.

$$2x^{2}-5x=7$$

Divide both sides by 2.

$$\frac{2x^{2}-5x}{2}=\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{5}{2}x=\frac{7}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\frac{7}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{7}{2}+\frac{25}{16}$$

Add \frac{7}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{9}{4} x-\frac{5}{4}=-\frac{9}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{7}{2} x=-1$$